How can I explicitly express the Ising Hamiltonian in matrix form?












2














I am reading this book about numerical methods in physics. It has the following question:




Consider the Ising Hamiltonian defined as following $$H=-sum_ {i=1}^{N-1}
sigma_i^x sigma_ {i+1} ^x + h sum_ {i=1} ^N sigma_i^z$$



Write a program that computes the $2^N times 2^N$ matrix for
different $N$




From quantum mechanics, I know that any operator can be expressed in matrix form as follows
$$H_{rs}dot{=}langle r |H|srangle$$
where $|irangle$ are any basis.



My question is what basis can I take for this Hamiltonian? Is there any other way to write this Hamiltonian in matrix form?










share|cite|improve this question




















  • 1




    Should be $H_{rs}$
    – InertialObserver
    5 hours ago






  • 1




    What does the subscript i mean? Are the Pauli matrixes for different is independent?
    – InertialObserver
    5 hours ago






  • 1




    Do you know how to write a program that computes the $2^Ntimes 2^N$ matrix for $H=sigma_i^x$ or $H=sigma_i^z$?
    – Norbert Schuch
    5 hours ago








  • 1




    I don't see why it's not just at simple as using the formula and looping over $i$
    – InertialObserver
    4 hours ago






  • 2




    Assuming $sigma$'s stand for spin-$frac{1}{2}$ pauli operators. Similar to what @NorbertSchuch is hinting at, one straightforward (perphaps numerically not very efficient way) is through using kronecker product of standard representations of pauli matrices.
    – Sunyam
    4 hours ago


















2














I am reading this book about numerical methods in physics. It has the following question:




Consider the Ising Hamiltonian defined as following $$H=-sum_ {i=1}^{N-1}
sigma_i^x sigma_ {i+1} ^x + h sum_ {i=1} ^N sigma_i^z$$



Write a program that computes the $2^N times 2^N$ matrix for
different $N$




From quantum mechanics, I know that any operator can be expressed in matrix form as follows
$$H_{rs}dot{=}langle r |H|srangle$$
where $|irangle$ are any basis.



My question is what basis can I take for this Hamiltonian? Is there any other way to write this Hamiltonian in matrix form?










share|cite|improve this question




















  • 1




    Should be $H_{rs}$
    – InertialObserver
    5 hours ago






  • 1




    What does the subscript i mean? Are the Pauli matrixes for different is independent?
    – InertialObserver
    5 hours ago






  • 1




    Do you know how to write a program that computes the $2^Ntimes 2^N$ matrix for $H=sigma_i^x$ or $H=sigma_i^z$?
    – Norbert Schuch
    5 hours ago








  • 1




    I don't see why it's not just at simple as using the formula and looping over $i$
    – InertialObserver
    4 hours ago






  • 2




    Assuming $sigma$'s stand for spin-$frac{1}{2}$ pauli operators. Similar to what @NorbertSchuch is hinting at, one straightforward (perphaps numerically not very efficient way) is through using kronecker product of standard representations of pauli matrices.
    – Sunyam
    4 hours ago
















2












2








2


1





I am reading this book about numerical methods in physics. It has the following question:




Consider the Ising Hamiltonian defined as following $$H=-sum_ {i=1}^{N-1}
sigma_i^x sigma_ {i+1} ^x + h sum_ {i=1} ^N sigma_i^z$$



Write a program that computes the $2^N times 2^N$ matrix for
different $N$




From quantum mechanics, I know that any operator can be expressed in matrix form as follows
$$H_{rs}dot{=}langle r |H|srangle$$
where $|irangle$ are any basis.



My question is what basis can I take for this Hamiltonian? Is there any other way to write this Hamiltonian in matrix form?










share|cite|improve this question















I am reading this book about numerical methods in physics. It has the following question:




Consider the Ising Hamiltonian defined as following $$H=-sum_ {i=1}^{N-1}
sigma_i^x sigma_ {i+1} ^x + h sum_ {i=1} ^N sigma_i^z$$



Write a program that computes the $2^N times 2^N$ matrix for
different $N$




From quantum mechanics, I know that any operator can be expressed in matrix form as follows
$$H_{rs}dot{=}langle r |H|srangle$$
where $|irangle$ are any basis.



My question is what basis can I take for this Hamiltonian? Is there any other way to write this Hamiltonian in matrix form?







condensed-matter quantum-spin computational-physics ising-model spin-chains






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 4 hours ago









Norbert Schuch

8,42122438




8,42122438










asked 5 hours ago









Luqman Saleem

249111




249111








  • 1




    Should be $H_{rs}$
    – InertialObserver
    5 hours ago






  • 1




    What does the subscript i mean? Are the Pauli matrixes for different is independent?
    – InertialObserver
    5 hours ago






  • 1




    Do you know how to write a program that computes the $2^Ntimes 2^N$ matrix for $H=sigma_i^x$ or $H=sigma_i^z$?
    – Norbert Schuch
    5 hours ago








  • 1




    I don't see why it's not just at simple as using the formula and looping over $i$
    – InertialObserver
    4 hours ago






  • 2




    Assuming $sigma$'s stand for spin-$frac{1}{2}$ pauli operators. Similar to what @NorbertSchuch is hinting at, one straightforward (perphaps numerically not very efficient way) is through using kronecker product of standard representations of pauli matrices.
    – Sunyam
    4 hours ago
















  • 1




    Should be $H_{rs}$
    – InertialObserver
    5 hours ago






  • 1




    What does the subscript i mean? Are the Pauli matrixes for different is independent?
    – InertialObserver
    5 hours ago






  • 1




    Do you know how to write a program that computes the $2^Ntimes 2^N$ matrix for $H=sigma_i^x$ or $H=sigma_i^z$?
    – Norbert Schuch
    5 hours ago








  • 1




    I don't see why it's not just at simple as using the formula and looping over $i$
    – InertialObserver
    4 hours ago






  • 2




    Assuming $sigma$'s stand for spin-$frac{1}{2}$ pauli operators. Similar to what @NorbertSchuch is hinting at, one straightforward (perphaps numerically not very efficient way) is through using kronecker product of standard representations of pauli matrices.
    – Sunyam
    4 hours ago










1




1




Should be $H_{rs}$
– InertialObserver
5 hours ago




Should be $H_{rs}$
– InertialObserver
5 hours ago




1




1




What does the subscript i mean? Are the Pauli matrixes for different is independent?
– InertialObserver
5 hours ago




What does the subscript i mean? Are the Pauli matrixes for different is independent?
– InertialObserver
5 hours ago




1




1




Do you know how to write a program that computes the $2^Ntimes 2^N$ matrix for $H=sigma_i^x$ or $H=sigma_i^z$?
– Norbert Schuch
5 hours ago






Do you know how to write a program that computes the $2^Ntimes 2^N$ matrix for $H=sigma_i^x$ or $H=sigma_i^z$?
– Norbert Schuch
5 hours ago






1




1




I don't see why it's not just at simple as using the formula and looping over $i$
– InertialObserver
4 hours ago




I don't see why it's not just at simple as using the formula and looping over $i$
– InertialObserver
4 hours ago




2




2




Assuming $sigma$'s stand for spin-$frac{1}{2}$ pauli operators. Similar to what @NorbertSchuch is hinting at, one straightforward (perphaps numerically not very efficient way) is through using kronecker product of standard representations of pauli matrices.
– Sunyam
4 hours ago






Assuming $sigma$'s stand for spin-$frac{1}{2}$ pauli operators. Similar to what @NorbertSchuch is hinting at, one straightforward (perphaps numerically not very efficient way) is through using kronecker product of standard representations of pauli matrices.
– Sunyam
4 hours ago












2 Answers
2






active

oldest

votes


















2














For completeness I'll summarize the answer here. After a fun conversation in the comments, we saw that it will be more illuminating to write



$$H=-sum_ {i=1}^{N-1}
sigma_i^x sigma_ {i+1} ^x + h sum_ {i=1} ^N sigma_i^z $$



as



$$ H=-sum_ {i=1}^{N-1}
sigma_i^x sigma_ {i+1} ^x + hleft( sum_ {i=1} ^{N-1} 1_1 otimes cdots 1_{i-1} otimes sigma_i^z otimes 1_{i+1} otimes cdots otimes 1_{N-1} right)
+ h ( 1_1 otimes cdots otimes 1_{N-1} otimes sigma^z_N) $$



where it is understood that (to prevent clutter)



$$ sigma_i^x sigma_{i+1}^x = 1_1 otimes cdotsotimes 1_{i-1} otimes sigma^x_i otimes sigma^x_{i+1} otimes 1_{i+2} otimes cdots otimes 1_{N} $$



and the subscripts are there to denote nothing more than the position of insertion (i.e. they are all $2times 2$ identity)



We then create a matrix from the direct product of matrices $A,B$ where $[A] = m times n$ and $[B] = p times q$ matrices



$$ mathbf {A} otimes mathbf {B} ={begin{bmatrix}a_{11}mathbf {B} &cdots &a_{1n}mathbf {B} \vdots &ddots &vdots \a_{m1}mathbf {B} &cdots &a_{mn}mathbf {B} end{bmatrix}}. $$



We apply this to the $i^{th}$ term and then loop over all $i$ to achieve the desired matrix.






share|cite|improve this answer























  • Your tensor product notation doesn't make sense: $sigma_N^z$ is $sigma^z$ acting on site $N$. You can't tensor this with $I_N$, which is $I$ acting on site $N$. You want to rather tensor it with $I$ on all other sites -- that's the definition of $sigma_N^z$.
    – Norbert Schuch
    4 hours ago










  • You're right. Thank you. I've made an edit but the notation is a bit sloppy still.
    – InertialObserver
    3 hours ago










  • Much better, but for consistency you should also add the identities in the first term!
    – Norbert Schuch
    3 hours ago










  • Will do, gonna do that now
    – InertialObserver
    3 hours ago






  • 1




    I see. Yes, that would screw up things dimensionally. Thanks a lot for clearing everything up.
    – InertialObserver
    3 hours ago



















2














$textbf{Quick hint :}$



Interpret spin operators (for chain of length-N) this way (spin operators act on tensor product space) :
$$sigma_{i}^{alpha} rightarrowunderset{N_{}^{text{th}}text{order direct/kronecker product of identity and pauli matrices}}{mathbb{I}_{}^{}otimescdotsotimesunderset{i_{}^{text{th}} text{position}}{sigma_{}^{alpha}}otimescdotsotimesmathbb{I}_{}^{}}$$
which makes (for $i < j$):



$$sigma_{i}^{alpha}sigma_{j}^{beta} stackrel{i < j}{rightarrow} mathbb{I}_{}^{}otimescdotsotimesunderset{i_{}^{text{th}} text{position}}{sigma_{}^{alpha}}otimescdotsotimesunderset{j_{}^{text{th}} text{position}}{sigma_{}^{beta}}otimes cdotsotimesmathbb{I}_{}^{}$$
with standard definition of direct/Kronecker product of matrices



Note also that Kronecker product is associative.



$mathbb{I}$ is $2 times 2$ identity matrix and $sigma_{}^{}$'s are standard Pauli matrices.






share|cite|improve this answer



















  • 2




    That was very helpful. Thank you.
    – Luqman Saleem
    4 hours ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














For completeness I'll summarize the answer here. After a fun conversation in the comments, we saw that it will be more illuminating to write



$$H=-sum_ {i=1}^{N-1}
sigma_i^x sigma_ {i+1} ^x + h sum_ {i=1} ^N sigma_i^z $$



as



$$ H=-sum_ {i=1}^{N-1}
sigma_i^x sigma_ {i+1} ^x + hleft( sum_ {i=1} ^{N-1} 1_1 otimes cdots 1_{i-1} otimes sigma_i^z otimes 1_{i+1} otimes cdots otimes 1_{N-1} right)
+ h ( 1_1 otimes cdots otimes 1_{N-1} otimes sigma^z_N) $$



where it is understood that (to prevent clutter)



$$ sigma_i^x sigma_{i+1}^x = 1_1 otimes cdotsotimes 1_{i-1} otimes sigma^x_i otimes sigma^x_{i+1} otimes 1_{i+2} otimes cdots otimes 1_{N} $$



and the subscripts are there to denote nothing more than the position of insertion (i.e. they are all $2times 2$ identity)



We then create a matrix from the direct product of matrices $A,B$ where $[A] = m times n$ and $[B] = p times q$ matrices



$$ mathbf {A} otimes mathbf {B} ={begin{bmatrix}a_{11}mathbf {B} &cdots &a_{1n}mathbf {B} \vdots &ddots &vdots \a_{m1}mathbf {B} &cdots &a_{mn}mathbf {B} end{bmatrix}}. $$



We apply this to the $i^{th}$ term and then loop over all $i$ to achieve the desired matrix.






share|cite|improve this answer























  • Your tensor product notation doesn't make sense: $sigma_N^z$ is $sigma^z$ acting on site $N$. You can't tensor this with $I_N$, which is $I$ acting on site $N$. You want to rather tensor it with $I$ on all other sites -- that's the definition of $sigma_N^z$.
    – Norbert Schuch
    4 hours ago










  • You're right. Thank you. I've made an edit but the notation is a bit sloppy still.
    – InertialObserver
    3 hours ago










  • Much better, but for consistency you should also add the identities in the first term!
    – Norbert Schuch
    3 hours ago










  • Will do, gonna do that now
    – InertialObserver
    3 hours ago






  • 1




    I see. Yes, that would screw up things dimensionally. Thanks a lot for clearing everything up.
    – InertialObserver
    3 hours ago
















2














For completeness I'll summarize the answer here. After a fun conversation in the comments, we saw that it will be more illuminating to write



$$H=-sum_ {i=1}^{N-1}
sigma_i^x sigma_ {i+1} ^x + h sum_ {i=1} ^N sigma_i^z $$



as



$$ H=-sum_ {i=1}^{N-1}
sigma_i^x sigma_ {i+1} ^x + hleft( sum_ {i=1} ^{N-1} 1_1 otimes cdots 1_{i-1} otimes sigma_i^z otimes 1_{i+1} otimes cdots otimes 1_{N-1} right)
+ h ( 1_1 otimes cdots otimes 1_{N-1} otimes sigma^z_N) $$



where it is understood that (to prevent clutter)



$$ sigma_i^x sigma_{i+1}^x = 1_1 otimes cdotsotimes 1_{i-1} otimes sigma^x_i otimes sigma^x_{i+1} otimes 1_{i+2} otimes cdots otimes 1_{N} $$



and the subscripts are there to denote nothing more than the position of insertion (i.e. they are all $2times 2$ identity)



We then create a matrix from the direct product of matrices $A,B$ where $[A] = m times n$ and $[B] = p times q$ matrices



$$ mathbf {A} otimes mathbf {B} ={begin{bmatrix}a_{11}mathbf {B} &cdots &a_{1n}mathbf {B} \vdots &ddots &vdots \a_{m1}mathbf {B} &cdots &a_{mn}mathbf {B} end{bmatrix}}. $$



We apply this to the $i^{th}$ term and then loop over all $i$ to achieve the desired matrix.






share|cite|improve this answer























  • Your tensor product notation doesn't make sense: $sigma_N^z$ is $sigma^z$ acting on site $N$. You can't tensor this with $I_N$, which is $I$ acting on site $N$. You want to rather tensor it with $I$ on all other sites -- that's the definition of $sigma_N^z$.
    – Norbert Schuch
    4 hours ago










  • You're right. Thank you. I've made an edit but the notation is a bit sloppy still.
    – InertialObserver
    3 hours ago










  • Much better, but for consistency you should also add the identities in the first term!
    – Norbert Schuch
    3 hours ago










  • Will do, gonna do that now
    – InertialObserver
    3 hours ago






  • 1




    I see. Yes, that would screw up things dimensionally. Thanks a lot for clearing everything up.
    – InertialObserver
    3 hours ago














2












2








2






For completeness I'll summarize the answer here. After a fun conversation in the comments, we saw that it will be more illuminating to write



$$H=-sum_ {i=1}^{N-1}
sigma_i^x sigma_ {i+1} ^x + h sum_ {i=1} ^N sigma_i^z $$



as



$$ H=-sum_ {i=1}^{N-1}
sigma_i^x sigma_ {i+1} ^x + hleft( sum_ {i=1} ^{N-1} 1_1 otimes cdots 1_{i-1} otimes sigma_i^z otimes 1_{i+1} otimes cdots otimes 1_{N-1} right)
+ h ( 1_1 otimes cdots otimes 1_{N-1} otimes sigma^z_N) $$



where it is understood that (to prevent clutter)



$$ sigma_i^x sigma_{i+1}^x = 1_1 otimes cdotsotimes 1_{i-1} otimes sigma^x_i otimes sigma^x_{i+1} otimes 1_{i+2} otimes cdots otimes 1_{N} $$



and the subscripts are there to denote nothing more than the position of insertion (i.e. they are all $2times 2$ identity)



We then create a matrix from the direct product of matrices $A,B$ where $[A] = m times n$ and $[B] = p times q$ matrices



$$ mathbf {A} otimes mathbf {B} ={begin{bmatrix}a_{11}mathbf {B} &cdots &a_{1n}mathbf {B} \vdots &ddots &vdots \a_{m1}mathbf {B} &cdots &a_{mn}mathbf {B} end{bmatrix}}. $$



We apply this to the $i^{th}$ term and then loop over all $i$ to achieve the desired matrix.






share|cite|improve this answer














For completeness I'll summarize the answer here. After a fun conversation in the comments, we saw that it will be more illuminating to write



$$H=-sum_ {i=1}^{N-1}
sigma_i^x sigma_ {i+1} ^x + h sum_ {i=1} ^N sigma_i^z $$



as



$$ H=-sum_ {i=1}^{N-1}
sigma_i^x sigma_ {i+1} ^x + hleft( sum_ {i=1} ^{N-1} 1_1 otimes cdots 1_{i-1} otimes sigma_i^z otimes 1_{i+1} otimes cdots otimes 1_{N-1} right)
+ h ( 1_1 otimes cdots otimes 1_{N-1} otimes sigma^z_N) $$



where it is understood that (to prevent clutter)



$$ sigma_i^x sigma_{i+1}^x = 1_1 otimes cdotsotimes 1_{i-1} otimes sigma^x_i otimes sigma^x_{i+1} otimes 1_{i+2} otimes cdots otimes 1_{N} $$



and the subscripts are there to denote nothing more than the position of insertion (i.e. they are all $2times 2$ identity)



We then create a matrix from the direct product of matrices $A,B$ where $[A] = m times n$ and $[B] = p times q$ matrices



$$ mathbf {A} otimes mathbf {B} ={begin{bmatrix}a_{11}mathbf {B} &cdots &a_{1n}mathbf {B} \vdots &ddots &vdots \a_{m1}mathbf {B} &cdots &a_{mn}mathbf {B} end{bmatrix}}. $$



We apply this to the $i^{th}$ term and then loop over all $i$ to achieve the desired matrix.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 3 hours ago

























answered 4 hours ago









InertialObserver

1,300517




1,300517












  • Your tensor product notation doesn't make sense: $sigma_N^z$ is $sigma^z$ acting on site $N$. You can't tensor this with $I_N$, which is $I$ acting on site $N$. You want to rather tensor it with $I$ on all other sites -- that's the definition of $sigma_N^z$.
    – Norbert Schuch
    4 hours ago










  • You're right. Thank you. I've made an edit but the notation is a bit sloppy still.
    – InertialObserver
    3 hours ago










  • Much better, but for consistency you should also add the identities in the first term!
    – Norbert Schuch
    3 hours ago










  • Will do, gonna do that now
    – InertialObserver
    3 hours ago






  • 1




    I see. Yes, that would screw up things dimensionally. Thanks a lot for clearing everything up.
    – InertialObserver
    3 hours ago


















  • Your tensor product notation doesn't make sense: $sigma_N^z$ is $sigma^z$ acting on site $N$. You can't tensor this with $I_N$, which is $I$ acting on site $N$. You want to rather tensor it with $I$ on all other sites -- that's the definition of $sigma_N^z$.
    – Norbert Schuch
    4 hours ago










  • You're right. Thank you. I've made an edit but the notation is a bit sloppy still.
    – InertialObserver
    3 hours ago










  • Much better, but for consistency you should also add the identities in the first term!
    – Norbert Schuch
    3 hours ago










  • Will do, gonna do that now
    – InertialObserver
    3 hours ago






  • 1




    I see. Yes, that would screw up things dimensionally. Thanks a lot for clearing everything up.
    – InertialObserver
    3 hours ago
















Your tensor product notation doesn't make sense: $sigma_N^z$ is $sigma^z$ acting on site $N$. You can't tensor this with $I_N$, which is $I$ acting on site $N$. You want to rather tensor it with $I$ on all other sites -- that's the definition of $sigma_N^z$.
– Norbert Schuch
4 hours ago




Your tensor product notation doesn't make sense: $sigma_N^z$ is $sigma^z$ acting on site $N$. You can't tensor this with $I_N$, which is $I$ acting on site $N$. You want to rather tensor it with $I$ on all other sites -- that's the definition of $sigma_N^z$.
– Norbert Schuch
4 hours ago












You're right. Thank you. I've made an edit but the notation is a bit sloppy still.
– InertialObserver
3 hours ago




You're right. Thank you. I've made an edit but the notation is a bit sloppy still.
– InertialObserver
3 hours ago












Much better, but for consistency you should also add the identities in the first term!
– Norbert Schuch
3 hours ago




Much better, but for consistency you should also add the identities in the first term!
– Norbert Schuch
3 hours ago












Will do, gonna do that now
– InertialObserver
3 hours ago




Will do, gonna do that now
– InertialObserver
3 hours ago




1




1




I see. Yes, that would screw up things dimensionally. Thanks a lot for clearing everything up.
– InertialObserver
3 hours ago




I see. Yes, that would screw up things dimensionally. Thanks a lot for clearing everything up.
– InertialObserver
3 hours ago











2














$textbf{Quick hint :}$



Interpret spin operators (for chain of length-N) this way (spin operators act on tensor product space) :
$$sigma_{i}^{alpha} rightarrowunderset{N_{}^{text{th}}text{order direct/kronecker product of identity and pauli matrices}}{mathbb{I}_{}^{}otimescdotsotimesunderset{i_{}^{text{th}} text{position}}{sigma_{}^{alpha}}otimescdotsotimesmathbb{I}_{}^{}}$$
which makes (for $i < j$):



$$sigma_{i}^{alpha}sigma_{j}^{beta} stackrel{i < j}{rightarrow} mathbb{I}_{}^{}otimescdotsotimesunderset{i_{}^{text{th}} text{position}}{sigma_{}^{alpha}}otimescdotsotimesunderset{j_{}^{text{th}} text{position}}{sigma_{}^{beta}}otimes cdotsotimesmathbb{I}_{}^{}$$
with standard definition of direct/Kronecker product of matrices



Note also that Kronecker product is associative.



$mathbb{I}$ is $2 times 2$ identity matrix and $sigma_{}^{}$'s are standard Pauli matrices.






share|cite|improve this answer



















  • 2




    That was very helpful. Thank you.
    – Luqman Saleem
    4 hours ago
















2














$textbf{Quick hint :}$



Interpret spin operators (for chain of length-N) this way (spin operators act on tensor product space) :
$$sigma_{i}^{alpha} rightarrowunderset{N_{}^{text{th}}text{order direct/kronecker product of identity and pauli matrices}}{mathbb{I}_{}^{}otimescdotsotimesunderset{i_{}^{text{th}} text{position}}{sigma_{}^{alpha}}otimescdotsotimesmathbb{I}_{}^{}}$$
which makes (for $i < j$):



$$sigma_{i}^{alpha}sigma_{j}^{beta} stackrel{i < j}{rightarrow} mathbb{I}_{}^{}otimescdotsotimesunderset{i_{}^{text{th}} text{position}}{sigma_{}^{alpha}}otimescdotsotimesunderset{j_{}^{text{th}} text{position}}{sigma_{}^{beta}}otimes cdotsotimesmathbb{I}_{}^{}$$
with standard definition of direct/Kronecker product of matrices



Note also that Kronecker product is associative.



$mathbb{I}$ is $2 times 2$ identity matrix and $sigma_{}^{}$'s are standard Pauli matrices.






share|cite|improve this answer



















  • 2




    That was very helpful. Thank you.
    – Luqman Saleem
    4 hours ago














2












2








2






$textbf{Quick hint :}$



Interpret spin operators (for chain of length-N) this way (spin operators act on tensor product space) :
$$sigma_{i}^{alpha} rightarrowunderset{N_{}^{text{th}}text{order direct/kronecker product of identity and pauli matrices}}{mathbb{I}_{}^{}otimescdotsotimesunderset{i_{}^{text{th}} text{position}}{sigma_{}^{alpha}}otimescdotsotimesmathbb{I}_{}^{}}$$
which makes (for $i < j$):



$$sigma_{i}^{alpha}sigma_{j}^{beta} stackrel{i < j}{rightarrow} mathbb{I}_{}^{}otimescdotsotimesunderset{i_{}^{text{th}} text{position}}{sigma_{}^{alpha}}otimescdotsotimesunderset{j_{}^{text{th}} text{position}}{sigma_{}^{beta}}otimes cdotsotimesmathbb{I}_{}^{}$$
with standard definition of direct/Kronecker product of matrices



Note also that Kronecker product is associative.



$mathbb{I}$ is $2 times 2$ identity matrix and $sigma_{}^{}$'s are standard Pauli matrices.






share|cite|improve this answer














$textbf{Quick hint :}$



Interpret spin operators (for chain of length-N) this way (spin operators act on tensor product space) :
$$sigma_{i}^{alpha} rightarrowunderset{N_{}^{text{th}}text{order direct/kronecker product of identity and pauli matrices}}{mathbb{I}_{}^{}otimescdotsotimesunderset{i_{}^{text{th}} text{position}}{sigma_{}^{alpha}}otimescdotsotimesmathbb{I}_{}^{}}$$
which makes (for $i < j$):



$$sigma_{i}^{alpha}sigma_{j}^{beta} stackrel{i < j}{rightarrow} mathbb{I}_{}^{}otimescdotsotimesunderset{i_{}^{text{th}} text{position}}{sigma_{}^{alpha}}otimescdotsotimesunderset{j_{}^{text{th}} text{position}}{sigma_{}^{beta}}otimes cdotsotimesmathbb{I}_{}^{}$$
with standard definition of direct/Kronecker product of matrices



Note also that Kronecker product is associative.



$mathbb{I}$ is $2 times 2$ identity matrix and $sigma_{}^{}$'s are standard Pauli matrices.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 3 hours ago

























answered 4 hours ago









Sunyam

6021311




6021311








  • 2




    That was very helpful. Thank you.
    – Luqman Saleem
    4 hours ago














  • 2




    That was very helpful. Thank you.
    – Luqman Saleem
    4 hours ago








2




2




That was very helpful. Thank you.
– Luqman Saleem
4 hours ago




That was very helpful. Thank you.
– Luqman Saleem
4 hours ago


















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